Two point charges of +2.0 \times 10^{-6}\text{ C} and +4.0 \times 10^{-6}\text{ C} are pla… — JAMB UTME

Two point charges of $+2.0 \times 10^{-6}\text{ C}$ and $+4.0 \times 10^{-6}\text{ C}$ are placed in a vacuum at a distance of $0.3\text{ m}$ apart. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}$ , what is the magnitude of the electrostatic force exerted between them in Newtons?

Answer: 0.8 N

Answer

The magnitude of the electrostatic force between the charges is 0.8 N.

According to Coulomb's Law, the force between two point charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them:

F = \frac{k q_1 q_2}{r^2}

. Substituting

q_1 = 2.0 \times 10^{-6}\text{ C}

q_2 = 4.0 \times 10^{-6}\text{ C}

r = 0.3\text{ m}

, and

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

into the formula gives

F = 0.8\text{ N}

Step-by-Step Solution

Identify known quantities from the problem statement

q_1 = 2.0 \times 10^{-6}\text{ C}

q_2 = 4.0 \times 10^{-6}\text{ C}

r = 0.3\text{ m}

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

Clear identification of parameters is required for substitution into Coulomb's Law.

Apply Coulomb's Law formula

F = \frac{k q_1 q_2}{r^2}

Coulomb's Law quantifies the electrostatic force between two stationary point charges.

Substitute values and perform arithmetic calculation

F = \frac{9.0 \times 10^9 \times (2.0 \times 10^{-6}) \times (4.0 \times 10^{-6})}{0.09} = \frac{0.072}{0.09} = 0.8\text{ N}

Squaring the separation distance

0.3\text{ m}

gives

0.09\text{ m}^2

, and evaluating the numerator gives

0.072\text{ N}\cdot\text{m}^2

Key Concept

Coulomb's Law

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